Finding a nullspace of a matrix - what should I do after finding equations? I am given the following matrix $A$ and I need to find a nullspace of this matrix.

$$A =
 \begin{pmatrix}
  2&4&12&-6&7 \\
  0&0&2&-3&-4 \\
  3&6&17&-10&7
 \end{pmatrix}$$

I have found a row reduced form of this matrix, which is:
$$A' =
 \begin{pmatrix}
  1&2&0&0&\frac{23}{10} \\
  0&0&1&0&\frac{13}{10} \\
  0&0&0&1&\frac{22}{10}  \end{pmatrix}$$
And then I used the formula $A'x=0$, which gave me:
$$A' =
 \begin{pmatrix}
  1&2&0&0&\frac{23}{10} \\
  0&0&1&0&\frac{13}{10} \\
  0&0&0&1&\frac{22}{10} 
 \end{pmatrix}
\begin{pmatrix}
  x_1 \\
  x_2 \\
  x_3  \\
  x_4 \\
  x_5
 \end{pmatrix}=
\begin{pmatrix}
  0 \\
  0 \\
  0  
 \end{pmatrix}$$
Hence I obtained the following system of linear equations:
$$\begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}$$
How should I proceed from this point?
Thanks!
 A: \begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}
$$
x_1=-2x_2-\dfrac{23}{10}x_5
$$
$$
x_3=-\dfrac{13}{10}x_5
$$
$$
x_4=-\dfrac{11}{5}x_5
$$
Therefore,basis of null space=
$$
\begin{pmatrix}
  -2 \\
  1 \\
  0  \\
  0 \\
  0
 \end{pmatrix},
\begin{pmatrix}
  -\dfrac{23}{10} \\
  0 \\
  -\dfrac{13}{10}  \\
  -\dfrac{11}{5} \\
  1
 \end{pmatrix}
$$
A: The null space of a matrix is basically a solution of the following: Ax = 0
x is a linear combination of all the independent matrices that satisfy the above equation. You have already multiplied the echelon form of A with x and have equated it to 0, so if any of the resulting equations is true, Ax = 0
We see that x5 is a variable in all the equations, so for the sake of simplicity, we take x5 = c. (a constant)
Thus, x3 = -10c/13 and x4 = -10c/22
x1, however, is still a problem because of the x2 for which we have no value. So we assign x2 another value, d (also a constant) and x1 = -2d - 10c/23
The two matrices which satisfy the equation Ax = 0 come out to be
(-2d  1d  0  0  0)^T and (-10c/23  0  -10c/13  -10c/22  c)^T
Let's call them y and z respectively.
If Ay = 0 and Az = 0 then k(Ay) + t(Az) = 0; where k and t are constants
Therefore, A (ky + tz) = 0
k * (-2d  1d  0  0  0)^T + t * (-10c/23  0 -10c/13  -10c/22  c)^T
is equivalent to (if we take d and c common in their matrices)
a * (-2  1  0  0  0)^T + b * (-10/23  0  -10/13  -10/22  1)^T
In short, Ax = 0 is satisfied by any linear combination of the following two matrices:
(-2  1  0  0  0)^T
(-10/23  0  -10/13  -10/22  1)^T
Together, they form the null space of A.
The null space of A belongs to: (-2  1  0  0  0)^T; (-10/23  0  -10/13  -10/22  1)^T
A: here is an almost algorithm to find a basis for the null space of a matrix $A:$
(a) row reduce $A,$ 
(b) identify the free and pivot b=variables. the variables corresponding to the non pivot columns, here they are $x_2$ and $x_4,$ are called the free variables and the rest are called pivot variables.
(c) you can set one of the free variables to one and the rest to zero. solve for the pivot variables. that gives you one solution in the null space; cycle through the free variables to find a basis for the null space.
i will show you how to  find one solution; you can find the other.
$x_2 = 1, x_4 = 0$  you can solve for $x_5$ first, giving you $x_5 = 0$ and then $x_3 = 0, x_1 = -2.$  so one vector in the basis is 
$$ \pmatrix{-2, &1, &0, &0, &0}^\top.$$
